Not acceleration the question is when two masses are released from rest determine the velocity with which m2 hits the ground

Since the system is released from rest and the pulley is frictionless, mechanical energy is conserved. The initial potential energy of $M_2$ is converted into kinetic energy of both masses and potential energy gained by $M_1$.

The initial total energy ($E_i$) is the potential energy of $M_2$ (assuming $M_1$ starts at a reference height of 0).

$$E_i = M_2 g h_2$$

The final total energy ($E_f$) is the potential energy of $M_1$ (which has moved up by $h_2$) plus the kinetic energy of both masses just before $M_2$ hits the ground.

$$E_f = M_1 g h_2 + \frac{1}{2} M_1 v_f^2 + \frac{1}{2} M_2 v_f^2$$

By conservation of energy, $E_i = E_f$.

$$M_2 g h_2 = M_1 g h_2 + \frac{1}{2} M_1 v_f^2 + \frac{1}{2} M_2 v_f^2$$

$$M_2 g h_2 - M_1 g h_2 = \frac{1}{2} (M_1 + M_2) v_f^2$$ $$g h_2 (M_2 - M_1) = \frac{1}{2} (M_1 + M_2) v_f^2$$ $$v_f^2 = \frac{2 g h_2 (M_2 - M_1)}{M_1 + M_2}$$ $$v_f = \sqrt{\frac{2 g h_2 (M_2 - M_1)}{M_1 + M_2}}$$

Given:
$M_1 = 1.5 \text{ kg}$
$M_2 = 4.5 \text{ kg}$
$h_2 = 2 \text{ m}$
$g = 9.81 \text{ m/s}^2$

$$v_f = \sqrt{\frac{2 \times 9.81 \text{ m/s}^2 \times 2 \text{ m} \times (4.5 \text{ kg} - 1.5 \text{ kg})}{1.5 \text{ kg} + 4.5 \text{ kg}}}$$ $$v_f = \sqrt{\frac{2 \times 9.81 \times 2 \times (3.0)}{6.0}} \text{ m/s}$$ $$v_f = \sqrt{\frac{117.72}{6.0}} \text{ m/s}$$ $$v_f = \sqrt{19.62} \text{ m/s}$$ $$v_f \approx 4.429 \text{ m/s}$$

The velocity with which $M_2$ hits the ground is approximately $4.43 \text{ m/s}$.